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          <h1 class="post-title" itemprop="name headline">PAT A1044</h1>
        

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        <p>Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken off the chain one by one. Once a diamond is off the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M$3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:</p>
<ol>
<li>Cut the chain between 4 and 6, and take off the diamonds from the position 1 to 5 (with values 3+2+1+5+4=15).</li>
<li>Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).</li>
<li>Cut before 8, and take off the diamonds from the position 7 to 8 (with values 8+7=15).<br>Now given the chain of diamond values and the amount that a customer has to pay, you are supposed to list all the paying options for the customer.</li>
</ol>
<p>If it is impossible to pay the exact amount, you must suggest solutions with minimum lost.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. For each case, the first line contains 2 numbers: N (&lt;=105), the total number of diamonds on the chain, and M (&lt;=108), the amount that the customer has to pay. Then the next line contains N positive numbers D1 … DN (Di&lt;=103 for all i=1, …, N) which are the values of the diamonds. All the numbers in a line are separated by a space.</p>
<p>Output Specification:</p>
<p>For each test case, print “i-j” in a line for each pair of i &lt;= j such that Di + … + Dj = M. Note that if there are more than one solution, all the solutions must be printed in increasing order of i.</p>
<p>If there is no solution, output “i-j” for pairs of i &lt;= j such that Di + … + Dj &gt; M with (Di + … + Dj - M) minimized. Again all the solutions must be printed in increasing order of i.</p>
<p>It is guaranteed that the total value of diamonds is sufficient to pay the given amount.</p>
<p>Sample Input 1:<br>16 15<br>3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13<br>Sample Output 1:<br>1-5<br>4-6<br>7-8<br>11-11<br>Sample Input 2:<br>5 13<br>2 4 5 7 9<br>Sample Output 2:<br>2-4<br>4-5</p>
<p>求等于S的子串：<br>思路，先令Sum表示1-i的值，这样做的好处是如果想知道3-5的和只要计算Sum[5]-Sum[2]就可以了<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">//#include &quot;math.h&quot;</span><br><span class="line">//#include &quot;string.h&quot;</span><br><span class="line">//#include &quot;algorithm&quot;</span><br><span class="line">//#include &quot;iostream&quot;</span><br><span class="line">//using namespace std;</span><br><span class="line">const int maxn = 100010;</span><br><span class="line">int sum[maxn];</span><br><span class="line">int n, S, nearS = 100000010;</span><br><span class="line">//upper_bound 函数返回在[L,R]内第一个大于x的位置</span><br><span class="line">int upper_bound(int L, int R, int x)&#123;</span><br><span class="line">    int left = L, right = R, mid;</span><br><span class="line">    while (left &lt; right) &#123;</span><br><span class="line">        mid = (left + right) / 2;</span><br><span class="line">        if (sum[mid] &gt; x) &#123;</span><br><span class="line">            right = mid;</span><br><span class="line">        &#125;else&#123;</span><br><span class="line">            left = mid + 1;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    return left;</span><br><span class="line">&#125;</span><br><span class="line">int main()&#123;</span><br><span class="line">    scanf(&quot;%d%d&quot;, &amp;n, &amp;S);</span><br><span class="line">    sum[0] = 0;</span><br><span class="line">    for (int i = 1; i &lt;= n ; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;sum[i]);</span><br><span class="line">        sum[i] += sum[i - 1];</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 1; i &lt;= n; i++) &#123;//遍历左端点</span><br><span class="line">        int j = upper_bound(i, n + 1, sum[i - 1] + S);//找右端点</span><br><span class="line">        if (sum[j - 1] - sum[i - 1] == S) &#123;</span><br><span class="line">            nearS = S;//查找成功，最接近S的就是S</span><br><span class="line">            break;</span><br><span class="line">        &#125;else if(j &lt;= n &amp;&amp; sum[j] - sum[i - 1] &lt; nearS)&#123;</span><br><span class="line">            nearS = sum[j] - sum[i - 1];//存在大于S的解并小鱼nearS</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 1; i &lt;= n; i++)&#123;</span><br><span class="line">        int j = upper_bound(i, n + 1, sum[i - 1] + S);//找右端点</span><br><span class="line">        if(S == nearS)&#123;</span><br><span class="line">            if (sum[j - 1] - sum[i - 1] == nearS) &#123;</span><br><span class="line">                printf(&quot;%d-%d\n&quot;, i, j - 1);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;else&#123;</span><br><span class="line">            if (sum[j] - sum[i - 1] == nearS) &#123;</span><br><span class="line">                printf(&quot;%d-%d\n&quot;, i, j);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">    &#125;</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<p>一会儿j-1一会儿j的原因是，upper_bound函数返回的是第一个大于x的位置<br>当sum[i-1]+s == sum[j] 时 upper_bound返回的值是left + 1,此时需要-1；<br>而找不到sum[i-1]+s == sum[j]时会返回一个最小的大于S的位置为left + 1，此时不需要-1。</p>

      
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